chapter one

3. Prove that if f is a real function on a measurable space X such that

x : f(x) ≥ r is measurable for every rational r, then f is measurable.

[proof]:For each real number α, there exists an descending sequence rn

of rational numbers such that limn→∞

rn = α. Moreover,we have

(α, +∞) =∞⋃

n=1

[rn, +∞).

Hence,

f−1((α, +∞)) =∞⋃

n=1

f−1([rn, +∞)).

Since sets f−1([rn, +∞)) are measurable for each n, the set f−1((α, +∞)) is

also measurable. Then f is measurable.

4. Let an and bn be sequences in [−∞, +∞], and prove the following

assertions:

(a)

lim supn→∞

(−an) = − lim infn→∞

an.

(b)

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn.

provided none of the sums is of the form ∞−∞.

(c) If an ≤ bn for all n, then

lim infn→∞

an ≤ lim infn→∞

bn.

Show by an example that strict inequality can hold in (b).

[proof]: (a) Since

supk≥n

(−ak) = − infk≥n

ak, n = 1, 2, · · · .

Therefore, let n → ∞, it obtains

limn→∞

supk≥n

(−ak) = − limn→∞

infk≥n

ak.

1

By the definations of the upper and the lower limits, that is

lim supn→∞

(−an) = − lim infn→∞

an.

(b) Since

supk≥n

(ak + bk) ≤ supk≥n

ak + supk≥n

bk, n = 1, 2, · · · .

Hence

limn→∞

supk≥n

(ak + bk) ≤ limn→∞

[supk≥n

ak + supk≥n

bk] = limn→∞

supk≥n

ak + limn→∞

supk≥n

bk.

By the definations of the upper and the lower limits, that is

lim supn→∞

(an + bn) ≤ lim supn→∞

an + lim supn→∞

bn.

example: we define

an = (−1)n, bn = (−1)n+1, n = 1, 2, · · · .

Then we have

an + bn = 0, n = 1, 2, · · · .

But

lim supn→∞

an = lim supn→∞

bn = 1.

(c)Because an ≤ bn for all n, then we have

infk≥n

(ak) ≤ infk≥n

bk, n = 1, 2, · · · .

By the definations of the lower limits, it follows

lim infn→∞

an ≤ lim infn→∞

bn.

5. (a)Supose f : X → [−∞, +∞] and g : X → [−∞, +∞] are measurable.

Prove that the sets

x : f(x) < g(x), f(x) = g(x)

are measurable.

(b)Prove that the set of points at which a sequence of measurable real-

valued functions converges (to a finite limit) is measurable.

2

[proof]: (a)Since for each rational r, the set

x : f(x) < r < g(x) = x : f(x) < r ∩ x : r < g(x)

is measurable. And

x : f(x) < g(x) =⋃r∈Q

x : f(x) < r < g(x)

Therefore the set x : f(x) < g(x) is measurable. Also beacause |f − g| is

measurable function and the set

x : f(x) = g(x) =∞⋂

n=1

x : |f(x) − g(x)| <1

n

is also measurable.

(b) Let fn(x) be the sequence of measurable real-valued functions, A

be the set of points at which the sequence of measurable real-valued functions

fn(x) converges (to a finite limit). Hence

A =∞⋂

r=1

∞⋃k=1

∞⋂n,m≥k

x : |fn(x) − fm(x)| <1

r.

Therefore, A is measurable.

7. Suppose that fn : X → [0, +∞] is measurable for n = 1, 2, 3, · · · , f1 ≥

f2 ≥ f3 ≥ · · · ≥ 0, fn(x) → f(x) as n → ∞,for every x ∈ Xand f1 ∈ L1(µ).

Prove that then

limn→∞

∫X

fndµ =

∫X

fdµ.

and show that this conclusion does not follow if the condition ”f1 ∈ L1(µ)”is

omitted.

[proof]: Define gn = f1−fn, n = 1, 2, · · ·, then gn is increasing measurable

on X for n = 1, 2, 3, · · ·. Moreover, gn(x) → f1(x)− f(x) as n → ∞, for every

x ∈ X. By the Lebesgue’s monotone convergence theorem, we have

limn→∞

∫X

gndµ =

∫X

(f1 − f)dµ.

Since f1 ∈ L1(µ), it shows

limn→∞

∫X

fndµ =

∫X

fdµ.

3

[counterexample]: Let X = (−∞, +∞), and we define

fn = χEn, En = [n, +∞), n = 1, 2, · · · .

8. Suppose that E is measurable subset of measure space (X, µ) with

µ(E) > 0; µ(X − E) > 0, then we can prove that the strict inequality in the

Fatou’s lemma can hold.

10. Suppose that µ(X) < ∞, fn is a sequence of bounded complex

measurable functions on X ,and fn → f uniformly on X. Proved that

limn→∞

∫X

fndµ =

∫X

fdµ,

and show that the hypothesis ”µ(X) < ∞”cannot be omitted.

[proof]: There exist Mn > 0, n = 1, 2, · · · such that |fn(x)| ≤ Mn for

every x ∈ X, n = 1, 2, · · ·. Since fn → f uniformly on X, there exists M > 0

such that

|fn(x)| ≤ M, |f(x)| ≤ M on X, n = 1, 2, · · · .

By the Lebesgue’s dominated convergence theorem, we have

limn→∞

∫X

fndµ =

∫X

fdµ.

[counterexample]: Let X = (−∞, +∞), and we define

fn =1

non X, n = 1, 2, · · · .

12. Suppose f ∈ L1(µ). Proved that to each ε > 0 there exists a δ > 0

such that∫

E|f |dµ < ε whenever µ(E) < δ.

[proof]: Since ∫X

|f |dµ = sup

∫X

sdµ,

the supremum being taken over all simple measurable functions s such that

0 ≤ s ≤ |f |. Hence, for each ε > 0, there exists a simple measurable function

s with 0 ≤ s ≤ |f | such that∫X

|f |dµ ≤

∫X

sdµ +ε

2.

Suppose that M > 0, satisfying 0 ≤ s(x) ≤ M on X, and let δ = ε

2M, we have∫

E

|f |dµ ≤

∫E

sdµ +ε

2≤ Mm(E) +

ε

2< ε

when m(E) < δ.

4

chapter two

1. Let fn be a sequence of real nonnegative functions on R1, and con-

sider the following four statements:

(a)If f1 and f2 are upper semicontinuous, then f1 + f2 is upper semicon-

tinuous.

(b)If f1 and f2 are lower semicontinuous, then f1 + f2 is lower semicon-

tinuous.

(c)If each fn is upper semicontinuous, then∞∑1

fn is upper semicontin-

uous.

(d)If each fn is lower semicontinuous, then∞∑1

fn is lower semicontinu-

ous.

Show that tree of these are true and that one is false. What happens if

the word ”nonnegative” is omitted? Is the truth of the statements affected if

R1 is replaced by a general topological space?

[proof]: (a) Because for every real number α,

x : f1(x) + f2(x) < α =⋃r∈Q

[x : f1(x) < r ∩ x : f2(x) < α − r]

and f1 and f2 are upper semicontinuous, therefore f1 + f2 is upper semicon-

tinuous.

(b) Because for every real number α,

x : f1(x) + f2(x) > α =⋃r∈Q

[x : f1(x) > r ∩ x : f2(x) > α − r]

and f1 and f2 are upper semicontinuous, therefore f1 + f2 is upper semicon-

tinuous.

(c) This conclusion is false. counterexample: for each n ∈ N, define

fn =

1; x = rn

0; others

1

which rn are the all rationals on R1. Then fn is a sequence of real non-

negative upper semicontinuous functions on R1, moreover we have

f(x) =∞∑1

fn(x) = χQ.

But f(x) is not upper semicontinuous.

(d) According to the conclusion of (b), it is easy to obtain.

If the word ”nonnegative” is omitted, (a) and (b) are still true but (c)

and (d) is not. The truth of the statements is not affected if R1 is replaced by

a general topological space.

2. Let f be an arbitrary complex function on R1, and define

φ(x, δ) = sup|f(s) − f(t)| : s, t ∈ (x − δ, x + δ)

φ(x) = infφ(x, δ) : δ > 0

Prove that φ is upper semicontinuous, that f is continuous at a point x if and

only if φ(x) = 0, and hence that the set of points of continuity of an arbitrary

complex function is a Gδ.

[proof]: It is obvious that

φ(x) = infφ(x,1

k) : k ∈ N ≥ 0

and

φ(x,1

k) ≥ φ(x,

1

k + 1), k ∈ N

For any α > 0, we will prove that the set x : φ(x) < α is open. Suppose

that φ(x) < α, then there exists a N > 0 such that

φ(x,1

k) < α, k > N.

Take y ∈ U(x, 1

2k), since U(y, 1

2k) ⊂ U(x, 1

k) and then

φ(y) ≤ φ(y,1

2k) ≤ φ(x,

1

k) < α

So we have y ∈ x : φ(x) < α. Therefore,

U(x,1

2k) ⊂ x : φ(x) < α

2

Thus, it show that x : φ(x) < α is open. Then it follows that φ is upper

semicontinuous. In the following, we first suppose that φ(x) = 0. for any

ε > 0, it exists δ > 0 such that

φ(x, δ) < ε

And so for any y ∈ (x − δ, x + δ) we have

|f(x) − f(y)| ≤ φ(x, δ) < ε

That is f is continuous at x. Next, provided that f is continuous at x. Then

for any ε > 0 it exists k ∈ N satifying

|f(x) − f(y)| ≤ε

2, ∀y ∈ (x −

1

k, x +

1

k)

and so φ(x, 1

k) ≤ ε. It shows

φ(x) = infφ(x,1

k) : k ∈ N = 0

Finally, since

x : φ(x) = 0 =

∞⋂n=1

x : φ(x) <1

n

then it shows that the set of points of continuity of an arbitrary complex

function is a Gδ.

3. Let X be a metric space, with metric ρ. For any nonempty E ⊂ X,

define

ρE(x) = infρ(x, y) : y ∈ E

Show that ρE is a uniformly continuous function on X. If A and B are disjoint

nonempty closed subsets of X, examine the relevance of the function

f(x) =ρA(x)

ρA(x) + ρB(x)

to Urysohn’s lemma.

[Proof]: Since for any x1, x2 ∈ X and y ∈ E, we have

ρ(x1, y) ≤ ρ(x1, x2) + ρ(x2, y)

Hence

ρE(x1) ≤ ρ(x1, x2) + ρE(x2)

3

and so

|ρE(x1) − ρE(x2)| ≤ ρ(x1, x2)

Therefore, it shows that ρE is a uniformly continuous function on X. In the

following, suppose that K ⊂ V ⊂ X, K is compact subset, and V is open

subset. Define

f(x) =ρV c(x)

ρV c(x) + ρK(x)

then 0 ≤ f ≤ 1 and f is continuous on X. Moreover, K ≺ f ≺ V .

4. Examine the proof of the Riesz theorem and prove the following two

statements:

(a) If E1 ⊂ V1 and E2 ⊂ V2, where V1 and V2 are disjoint open sets, then

µ(E1 ∪ E2) = µ(E1) + µ(E2),

even if E1 and E2 are not in M.

(b) If E ∈ MF , then E = N ∪ K1 ∪ K2 ∪ · · ·, where Ki is a disjoint

countable collection of compact sets and µ(N) = 0.

[proof]: (a) By the proof of the Riesz theorem, there exists a Gδ set G

such that

E1 ∪ E2 ⊂ G and µ(E1 ∪ E2) = µ(G).

Set G1 = V1 ∩ G, G2 = V2 ∩ G, then E1 ⊂ G1, E2 ⊂ G2 and G1 ∩ G2 = ∅.

Therefore

µ(E1)+µ(E2) ≤ µ(G1)+µ(G2) = µ(G1∪G2) = µ(G∩(V1∪V2)) ≤ µ(G) = µ(E1∪E2).

And since

µ(E1 ∪ E2) ≤ µ(E1) + µ(E2),

It shows

µ(E1 ∪ E2) = µ(E1) + µ(E2).

(b) Since E ∈ MF , there exist a sequence of disjoint compact subsets

Kn such that

Kn ⊂ E −

n−1⋃i=0

Ki, m(E −

n−1⋃i=0

Ki) < m(Kn) +1

n, n = 1, 2, · · · ,

which we order K0 = ∅.

4

Set K =∞⋃

n=1

Kn. Since m(E) < +∞, we obtain that

m(E −n⋃

i=1

Ki) <1

n, n = 1, 2, · · · .

Whence we have

m(E − K) ≤ m(E −

n⋃i=1

Ki) <1

n, n = 1, 2, · · · .

Set N = E − K, then it shows that m(N) = 0 and E = K ∪ N .

5. Let E be Cantor’s familiar ”middle thirds” set. Show that m(E) = 0.

even though E and R1 have the same cardinality.

[proof]: According to the construction of Cantor’s set, it shows that the

set E is closed set and

m(E) = m([0, 1]) −

∞∑n=1

2n−1

3n= 0.

11. Let µ be a regular Borel measure on a compact Hausdorff space

X; assume µ(X) = 1. prove that there is a compact set K ⊂ X such that

µ(K) = 1 but µ(H) < 1 for every proper compact subset H of K.

[Proof]: Define the set

Ω = Kα : Kα ⊂ X, Kα is compact set with µ(kα) = 1

then X ∈ Ω. Order

K =⋂

Kα∈Ω

Kα

Thus K is compact subset. Assume that V ⊂ X is an open set with K ⊂ V .

Since X is compact, V c is also compact. Again since

V c ⊂⋃

Kα∈Ω

Kcα

then there exist finite sets Kα1, · · · , Kαn

such that

V c ⊂n⋃

k=1

Kcα

k

5

Because µ(X) = 1 = µ(Kα),

µ(Kcα

k

) = 0, k = 1, 2, · · · , n

and so µ(V c) = 0. We therefore have µ(V ) = 1. Since µ is regular, hence it

shows that µ(K) = 1. By the construction of K, it follows that µ(H) < 1

for any proper compact subset of K. If U ⊂ X is open set with µ(U) = 0, it

follows µ(U c) = 1. U c is compact subset since X is compact, and so K ⊂ U c

by the construction of K. Hence, U ⊂ Kc. It thus shows that Kc is the largest

open set in X whose measure is 0.

12. Show that every compact subset of R1 is the support of a Borel

measure.

[Proof]: Assume that µ is the Lebesgue measure on R1 and K is a com-

pact subset of R1. Define

φ(f) =

∫K

fdµ; f ∈ Cc(R1).

Then φ is a positive linear functional on Cc(R1). By Riesz representation

theorem, there exists a σ-algebra M in R1 which contains all Borel sets in R

1,

and there exists a unique positive measure σ on M such that

φ(f) =

∫R1

fdσ, f ∈ Cc(R1).

By theorem2.14, it shows

σ(K) = infφ(f) : K ≺ f = µ(K)

σ(X) = supφ(f) : f ≺ X ≤ µ(K)

and so σ(X) = σ(K) = µ(K) < +∞. Therefore, it follows for any measurable

set A ⊂ X,

σ(A) = σ(A ∩ K).

That is, K is the support of σ.

14. Let f be a real-valued Lebesgue measurable function on Rk. Prove

that there exist Borel functions g and h such that g(x) = h(x) a.e.[m] and

g(x) ≤ f(x) ≤ h(x) for every x ∈ Rk.

6

[Proof]: Firstly, assume that 1 > f ≥ 0. By the proof of Theorem 2.24

in text, we have

f(x) =∞∑

n=1

tn(x), ∀x ∈ Rk

and 2ntn is the characteristic function of a Lebesgue measurable set Tn ⊂

Rk(n ∈ N). According to the construction of Lebesgue measurable set, there

exist Borel measurable sets Fn, En(n ∈ N) satisfying

Fn ⊂ Tn ⊂ En; m(En − Fn) = 0; n ∈ N.

Define

g =∞∑

n=1

2−nχFn; h =

∞∑n=1

2−nχEn

then g, h are Borel measurable functions on Rk and g(x) ≤ f(x) ≤ h(x).

Moreover, g(x) = h(x) a.e.[m].

Secondly, it is easy that the conclusion is correct for 0 ≤ f < M , and

hence it is also for bounded real-valued Lebesgue measurable function on Rk.

Finally, if f be a real-valued Lebesgue measurable function on Rk and if

Bn = x : |f(x)| ≤ n, n = 1, 2, · · ·

then we have

χBn(x)f(x) → f(x), as n → ∞.

According to the conclusion above, there exist Borel functions gn and hn such

that gn(x) = hn(x) a.e.[m] and gn(x) ≤ χBn(x)f(x) ≤ hn(x) for every x ∈ R

k.

Take

g = lim sup gn; h = lim inf hn

then g(x) = h(x) a.e.[m] and g(x) ≤ f(x) ≤ h(x) for every x ∈ Rk.

15. It is easy to guess the limits of∫ n

0

(1 −x

n)ne

x

2 dx and

∫ n

0

(1 +x

n)ne−2xdx

as n → ∞. Prove that your guesses are correct.

[proof]: Define

fn(x) = χ[0,n](1 −x

n)ne

x

2 ; gn(x) = χ[0,n](1 +x

n)ne−2x; n ∈ N

7

Then it is easy that

|fn(x)| ≤ e−x

2 , and fn(x) → e−x

2 ;

|gn(x)| ≤ e−x, and fn(x) → e−x.

By the Lebesgue’s Dominated Convergence Theorem, it shows

limn→∞

∫ n

0

(1 −x

n)ne

x

2 dx = limn→∞

∫ ∞

0

fn(x)dx =

∫ ∞

0

e−x

2 dx = 2

limn→∞

∫ n

0

(1 +x

n)ne−2xdx = lim

n→∞

∫ ∞

0

gn(x)dx =

∫ ∞

0

e−xdx = 1

17. Define the distance between points (x1, y1) and (x2, y2) in the plane

to be

|y1 − y2| if x1 = x2; 1 + |y1 − y2| if x1 6= x2.

show that this is indeed a metric, and that the resulting metric space X is

locally compact.

If f ∈ Cc(X), let x1, x2, · · · , xn be those values of x for which f(x, y) 6= 0

for at least one y (there are only finitely many such x!), and define

Λf =

n∑j=1

∫+∞

−∞

f(xj, y)dy

Let µ be the measure associated with this Λ by Theorem2.14. If E is the

x-axis, show that µ(E) = ∞ although µ(K) = 0 for every compact K ⊂ E.

[proof]:Order

Pi = (xi, yi), i = 0, 1, 2, 3, · · · .

d(P1, P2) =

|y1 − y2| if x1 = x2;

1 + |y1 − y2| if x1 6= x2

Since

d(P1, P2) ≤

|y1 − y3| + |y2 − y3| if x1 = x2;

1 + |y1 − y3| + |y2 − y3| if x1 6= x2

Therefore,

d(P1, P2) ≤ d(P1, P3) + d(P2, P3).

That is d is metric. Take X = (R2, d). Then

Pn → P ⇔ ∃ N > 0, such that if n > N, xn = x; and|yn − y| → 0.

8

and for any 0 < δ < 1,

U(P0, δ) = P ∈ X : d(P0, P ) < δ = P ∈ X : x = x0, and |y − y0| < δ

U(P0, δ) = P ∈ X : d(P0, P ) ≤ δ = P ∈ X : x = x0, and |y − y0| ≤ δ

Moreover, U(P0, δ) is compact subset of X. Thus X is locally compact Haus-

dorff space.

In the following, we order the set for f ∈ Cc(X)

Ωf = x ∈ R : ∃ y ∈ R such that f(x, y) 6= 0

Let K be the support of f , K is compact. Then there exists finite P1, P2, · · · , Pn ∈

K for any 0 < δ < 1 such that

K ⊂

n⋃j=1

U(Pj, δ)

and so Ωf = x1, x2, · · · , xn. Since

Λf =

n∑j=1

∫+∞

−∞

f(xj, y)dy

it is easy that Λ is a positive linear functional on Cc(X). There thus exist a

measure µ associated with this Λ by Theorem2.14. Let E be the x-axis and

0 < δ < 1. Define

V =⋃

Pi∈E

U(Pi, δ); where Pi = (xi, 0);

it shows V is open and E ⊂ V . For any finite elements P1, P2, · · · , Pn ∈ E, the

set

F =

n⋃j=1

U(Pj,δ

2)

is compact with F ⊂ V . By the Urysohn’s lemma, there exists f ∈ Cc(X)

satisfying F ≺ f ≺ V . So we have

µ(V ) ≥ Λf ≥ nδ.

Hence, it shows µ(V ) = ∞ as n → ∞. Therefore, µ(E) = ∞. Assume that

K ⊂ E is compact, then it is obvious that K is finite set. Take

K = P1, P2, · · · , Pm

9

For any 0 < ε < 1, define

G =

m⋃j=1

U(Pj, ε)

then g is open and K ⊂ G. By the Urysohn’s lemma, there exists g ∈ Cc(X)

satisfying K ≺ g ≺ G. Since

µ(K) ≤ Λg ≤ 2ε

whence µ(K) = 0.

21. If X is compact and f : X → (−∞, +∞) is upper semicontinuous,

prove that f attains its maximum at some point of X.

[proof]: for every t ∈ f(X), define

Et = x ∈ X : f(x) < t

Assume that f does not attain its maximum at some point of X. Then it

shows X =⋃

t∈f(X)

Et. Since f is continuous, Et is open sets. Again since X is

compact, there exists finite ti(i = 1, 2, · · · , n) ∈ f(X) such that X =n⋃

i=1

Eti .

Take

t0 = maxt1, t2, · · · , tn

then we have

f(x) < t0, ∀x ∈ X.

But it is in contradiction with t0 ∈ f(X). It therefore is proved that f attains

its maximum at some point of X.

10

chapter three

1. Prove that the supremum of any collection of convex functions on (a, b)

is convex on (a, b) and that pointwise limits of sequences of convex functions

are convex. What can you say about upper and lower limits of sequences of

convex functions?

[Proof]: Suppose that α ≥ 0, β ≥ 0 and α + β = 1. Let fii∈I be any

collection of convex functions on (a, b). for any x, y ∈ (a, b), we have

fi(αx+ βy) ≤ αfi(x) + βfi(y), ∀i ∈ I.

It prove that

supi∈I

fi(αx+ βy) ≤ α supi∈I

fi(x) + β supi∈I

fi(y).

So supi∈I

fi is convex.

Assume I = N and f is pointwise limits of sequences of convex functions fi.

according to the properties of limit, we have

limi→∞

fi(αx+ βy) ≤ α limi→∞

fi(x) + β limi→∞

fi(y)

That is f(αx+ βy) ≤ αf(x) + βf(y) and so f is convex.

By the definition of upper limits and the conclusion above, it is easy that the

upper limits of sequences of convex functions is convex. But the lower limits

of sequences of convex functions is false.

2. If φ is convex on (a, b) and if ψ is convex and nondecreasing on the

range of φ, prove that ψ φ is convex on (a, b). For φ > 0, show that the

convexity of lnφ implies the convexity of φ , but not vice versa.

[Proof]: Assume that

x, y ∈ (a, b);α, β ≥ 0and α + β = 1.

Then

ψ φ(αx+ βy) ≤ ψ(αφ(x) + βφ(y)) ≤ αψ φ(x) + βψ φ(y).

1

and so ψ φ is convex on (a, b). Since et is nondecreasing convex on R and

φ = eln φ, it shows from the conclusion above that the convexity of lnφ implies

the convexity of φ for φ > 0.

[counterexample]: φ(x) = x2 is convex on (0,∞) but lnφ(x) = 2 lnx is

not convex. Moreover, for φ > 0, it can show that the convexity of logc φ(c > 1)

implies the convexity of φ.

3. Assume that φ is a continuous real function on (a, b) such that

φ(x + y

2) ≤

1

2φ(x) +

1

2φ(y)

for all x and y ∈ (a, b). Prove that φ is convex.

[Proof]: According to the definition of convex function,we only need to

prove the case 0 < λ < 1

2. Take

E = k

2n| k = 1, 2, · · · , 2n − 1; n ∈ N.

For n = 1, it is obvious that

φ(x + y

2) ≤

1

2φ(x) +

1

2φ(y)

for all x and y ∈ (a, b).

For n = 2, suppose that λk = k

4, k = 1.

φ(λ1x+ (1 − λ1)y) = φ(1

2(x+y

2+ y))

≤ 1

2φ(x+y

2) + 1

2φ(y)

≤ 1

2[12φ(x) + 1

2φ(y)] + 1

2φ(y)

= 1

4φ(x) + 3

4φ(y)

= λ1φ(x) + (1 − λ1)φ(y)

Assume that the case

λ =k

2n−1, k = 1, 2, · · · , 2n−1 − 1

is correct. For any λ ∈ E with λ = k

2n; 1 ≤ k ≤ 2n−1, n ≥ 2, and for all x and

y ∈ (a, b), it shows that

(i)

φ(λx+ (1 − λ)y) = φ(x+ y

2) ≤

1

2φ(x) +

1

2φ(y) = λφ(x) + (1 − λ)φ(y)

2

when k = 2n−1.

(ii)for 1 ≤ k < 2n−1,

λx + (1 − λy) =k

2nx + (1 −

k

2n)y =

1

2(k

2n−1x+ (1 −

k

2n−1)y + y)

Since k

2n−1x+ (1 − k

2n−1 )y ∈ (a, b) and so

φ(λx + (1 − λ)y) ≤ 1

2φ( k

2n−1x + (1 − k

2n−1 )y) + 1

2φ(y)

≤ k

2nφ(x) + 1

2(1 − k

2n−1 )φ(y) + 1

2φ(y)

= λφ(x) + (1 − λ)φ(y)

For any 0 < λ < 1, there exists a sequence λn ⊂ E with limn→∞

λn = λ.

Thusφ(λx+ (1 − λ)y) = lim

n→∞φ(λnx + (1 − λn)y)

≤ limn→∞

(λnφ(x) + (1 − λn)φ(y))

= λφ(x) + (1 − λ)φ(y)

Therefore, it shows that φ is convex on (a, b).

4. Suppose f is a complex measurable function on X, µ is a positive

measure on X, and

φ(p) =

∫X

|f |pdµ = ‖f‖pp (0 < p <∞).

Let E = p : φ(p) <∞, Assume ‖f‖∞ > 0.

(a) If r < p < s, r ∈ E, and s ∈ E, Prove that p ∈ E.

(b) Prove that lnφ is convex in the interior of E and that φ is continuous

on E.

(c) By (a), E is connected. Is E necessarily open? losed? Can E consist

of a single point? Can E be any connected subset of (0,∞)?

(d) If r < p < s, prove that ‖f‖p ≤ max(‖f‖r, ‖f‖s). Show that this

implies the inclusion Lr(µ) ∩ Ls(µ) ⊂ Lp(µ).

(e) Assume that ‖f‖r <∞ for some r <∞ and prove that

‖f‖p → ‖f‖∞ as p→ ∞.

[Proof]: (a) If r < p < s, there exist 0 < α < 1 and 0 < β < 1 such that

α + β = 1 and p = αr + βs. By Holder’s inequality, it follows that

‖f‖pp = φ(p) =

∫X

|f |pdµ =

∫X

|f |αr+βsdµ ≤

∫X

|f |rdµα·

∫X

|f |sdµβ = ‖f‖αrr ·‖f‖βs

s .

3

As r ∈ E and s ∈ E, it prove that p ∈ E.

(b) By (a), it reduces that E is convex. Moreover, E is connected. Since

E ⊂ (0,∞), the interior of E is an interval and denoted by (a, b). For any s, t

in the interior of E and 0 < α < 1 and 0 < β < 1, it follows from (a) that

φ(αs+ βt) ≤ φ(s)α · φ(t)β.

So it proves

lnφ(αs+ βt) ≤ ln[φ(s)α · φ(t)β] = α lnφ(s) + β lnφ(t).

That is, lnφ is convex in the interior of E. Moreover, lnφ is continuous on

(a, b). Thus, φ is also continuous on (a, b). Provided that a ∈ E. For any

decreasing sequence sn of (a, b) with limn→∞

sn = a, it is easy to show that

limn→∞

|f(x)|sn = |f(x)|a, ∀x ∈ X.

Since |f(x)|sn ≤ max |f(x)|a, |f(x)|s1, ∀x ∈ X and |f(x)|a, |f(x)|s1 ∈ L1(µ),

by the Lebesgue control convergence theorem, it follows that

limn→∞

φ(sn) = φ(a).

Similarly, the other cases can be proved. Hence φ is continuous on E.

(c) By (a), it shows that E is connected and convex. E can be any interval

on (0,∞). For examples: take X = (0,∞) and µ is the Lebesgue measure on

X.

(i) For E = (a, b), (0 < a < b), take

f(x) =

x−

1

b , x ∈ (0, 1);

x−1

a , x ∈ [1,∞)

(ii)For E = [a, b], (0 < a < b), take

f(x) =

x, x ∈ (0, 1);

x−1

a−1, x ∈ [1,∞)

(iii)For E = [a, b), (0 < a < b), take

f(x) =

x−

1

b , x ∈ (0, 1);

x−1

a−1, x ∈ [1,∞)

4

(iv)For E = (a, b], (0 < a < b), take

f(x) =

x−

1

b+1 , x ∈ (0, 1);

x−1

a , x ∈ [1,∞)

(v)For E = ∅, take f ≡ 1.

(d) If r < p < s,by (a), there exist 0 < α < 1 and 0 < β < 1 such that

α + β = 1 and p = αr + βs. And it is easy to know

φ(p) = φ(αr + βs) ≤ φ(r)α · φ(s)β ≤ maxφ(r), φ(s).

Thus, ‖f‖p ≤ max ‖f‖r, ‖f‖s. Moreover, it shows that Lr(µ)∩Ls(µ) ⊂ Lp(µ).

(e)Firstly since ‖f‖∞ > 0, it shows that ‖f‖p > 0 for any 0 < p < ∞.

secondly, assume that ‖f‖∞ = +∞. Hence,µ(x ∈ x : |f(x)| > 2) > 0. For

any p > r,

0 < 2pµ(x ∈ x : |f(x)| > 2) ≤

∫X

|f |pdµ = ‖f‖pp.

Therefore, limn→∞

‖f‖p = +∞.

Thirdly, assume that ‖f‖∞ = 1 and ‖f‖r > 0. Without generality, assume

that |f(x)| ≤ 1, ∀x ∈ X. Thus For any p > r, it shows that

‖f‖pp ≤ ‖f‖r

r.

That is ‖f‖p ≤ ‖f‖r

p

r . And so

limp→∞

‖f‖p = 1 = ‖f‖∞.

The general case can reduce to the case ‖f‖∞ = 1 and ‖f‖r > 0. This is

completed the proof.

5. Assume, in addition to the hypotheses of Exercise 4, that

µ(X) = 1.

(a) Prove that ‖f‖r ≤ ‖f‖s, if 0 < r < s ≤ ∞.

(b) Under what conditions does it happen that 0 < r < s ≤ ∞ and ‖f‖r =

‖f‖s <∞?

(c) Prove that Lr(µ) ⊃ Ls(µ) if 0 < r < s. Under what conditions do these

5

two spaces contain the same functions?

(d) Assume that ‖f‖r <∞ for some r > 0, and prove that

limp→0

‖f‖p = exp

∫X

log |f |dµ

if exp−∞ is defined to be 0.

[Proof]: (a) When s = ∞, it is easy that∫

X

|f |rdµ ≤ ‖f‖r∞µ(X)

and so ‖f‖r ≤ ‖f‖∞. If s < ∞, φ(x) = xs

r is convex on (0,∞) and by the

Jensen’s Inequality, it shows∫X

|f |sdµ =

∫X

(|f |r)s

rdµ ≥ (

∫X

(|f |r)dµ)s

r

and whence ‖f‖s ≥ ‖f‖r.

(b) Assume that s = ∞, and ‖f‖r = ‖f‖∞ < ∞(0 < r < s). Then for

any 1 < p <∞, it shows by (a) that

‖f‖1 = ‖f‖p = ‖f‖q <∞

and so f is constant almost everywhere by the Holder’s inequality. Secondly,

assume 0 < r < s <∞. Take α = sr> 1, then by the Holder’s inequality

∫X

|f |r · 1dµ ≤ (

∫X

|f |sdµ)r

s = ‖f‖rs = ‖f‖r

r

Therefore, |f |r = constant a.e. on X. So it happen that 0 < r < s ≤ ∞ and

‖f‖r = ‖f‖s <∞ if and only if f = constant a.e. on X.

(c)If 0 < r < s, then by (a) it is easy to prove Ls(µ) ⊂ Lr(µ). Moreover,

we will prove the following:

Lr(µ) = Ls(µ) ⇔ there exist finite pairwise disjoint measurable sets at most in

Ω = E ⊂ X : µ(E) > 0.

Firstly, assume that there are infinte pairwise disjoint members in Ω. Then it

exists a sequence En ⊂ X of pairwise disjoint measurable sets such that

0 < µ(En) < 1, n = 1, 2, · · · .

6

Take 1

s< t < 1

r, and define

f =

∞∑n=1

n−tχEn

Then ∫X|f |sdµ =

∫X

∑∞

n=1n−tsχEn

dµ

=∑∞

n=1n−tsµEn

≤∑∞

n=1n−ts

Since st > 1, it has∑∞

n=1n−t < +∞ and so f ∈ Ls(µ). But

∫X|f |rdµ =

∫X

∑∞

n=1n−trχEn

dµ

=∑∞

n=1n−trµEn

≥∑∞

n=1n−tr · n−α

=∑∞

n=1n−(tr+α)

which α = 1−tr2

. Since 0 < α + tr < 1, it shows that∑∞

n=1n−(tr+α) = +∞.

Hence, f /∈ Lr(µ).

Next, we assume that there exist finite pairwise disjoint measurable sets at

most in Ω. Provided that f ∈ Lr(µ), then there exists an increasing seqnence

hn of simple measurable functions such that

0 ≤ hn(x) ≤ hn+1(x) ≤ · · · ≤ |f(x)|, forallx ∈ X

and

hn(x) → |f(x)|, n→ ∞, ∀x ∈ X.

Put

t1(x) = h1(x); tn(x) = hn(x) − hn−1(x), n = 2, 3, · · ·

then 2ntn is the characteristic function of a measurable set Tn ⊂ X, and Tn

are pairwise disjoint. Moreover,

f(x) =∞∑

n=1

tn(x), ∀x ∈ X.

According to the assumption, there are finite members of Tn with nonzero

measure at most. Thus f equal to a simple function almost everywhere and

so f ∈ Ls(µ). Therefore, Lr(µ) = Ls(µ).

7

(d) Assume that ‖f‖r <∞ for some r > 0, then

‖f‖p ≤ ‖f‖r < +∞; ‖f‖p ≥ e∫

Xln |f |dµ; 0 < p < r.

and ‖f‖p is increasing relative to p. Therefore, the limit limp→0

‖f‖p = c exists.

If c=0, it is easy that e∫

Xln |f |dµ = 0. In the following, we assume c > 0. Since

‖f‖r < +∞, we may suppose that f is finite everywhere. Let

E = [|f | > 1]; F = [|f | < 1]; G = [|f | = 1]

So we have

‖f‖p = (∫

X|f |pdµ)

1

p

= (∫

E(|f |p − 1 + 1)dµ+

∫F(|f |p − 1 + 1)dµ+

∫G(|f |p − 1 + 1)dµ)

1

p

= (1 +∫

E(|f |p − 1)dµ+

∫F(|f |p − 1)dµ)

1

p

Take α > 1; 0 < β < 1, because

limx→1

x− 1

ln x= 1; |f(x)|p → 1, p→ 0

there exists δ > 0 such that

|f(x)|p − 1 < α ln |f(x)|p, ∀x ∈ E; |f(x)|p − 1 < β ln |f(x)|p, ∀x ∈ F

when p < δ. Thus

‖f‖p ≤ (1 +∫

Eα ln |f |pdµ+

∫Fβ ln |f |pdµ)

1

p

= (1 + p(α∫

Eln |f |dµ+ β

∫F

ln |f |dµ)1

p

Then

0 < c = limp→0

‖f‖p ≤ eα∫

Eln |f |dµ+β

∫

Fln |f |dµ

Let

α→ 1+; β → 1−

then

c = limp→0

‖f‖p ≤ e∫

Eln |f |dµ+

∫

Fln |f |dµ = e

∫

Xln |f |dµ.

Hence

limp→0

‖f‖p = e∫

Xln |f |dµ.

8

11. Suppose µ(Ω) = 1, and suppose f and g are positive measurable

functions on Ω such that fg ≥ 1. Prove that∫

Ω

fdµ ·

∫Ω

gdµ ≥ 1.

[Proof]: Since f and g are positive measurable functions on Ω such that

fg ≥ 1, f1

2 and g1

2 are positive measurable functions on Ω such that (fg)1

2 ≥ 1.

By Holder’s inequality, it follows that

1 ≤ (

∫ω

(fg)1

2dµ)2 ≤

∫Ω

fdµ ·

∫Ω

gdµ.

12. Suppose µ(Ω) = 1 and h : Ω → [0,∞] is measurable. If

A =

∫Ω

hdµ,

prove that√

1 + A2 ≤

∫Ω

√1 + h2dµ ≤ 1 + A.

[Proof]: It is obvious correct when A = ∞ or 0 and so we assume 0 <

A < +∞ and h : Ω → (0,∞). Since φ(x) =√

1 + x2 is convex on (0,∞), it

shows by the Jensen’s Inequality that

√1 + A2 ≤

∫Ω

√1 + h2dµ.

Because√

1 + h2 ≤ 1 + h,∫

Ω

√1 + h2dµ ≤ 1 + A.

This complete the proof.

14. Suppose 1 < p < ∞, f ∈ Lp = Lp((0,∞)), relative to Lebesgue

measure and

F (x) =1

x

∫ x

0

f(t)dt (0 < x <∞).

(a) Prove Hardy’s inequality

‖F‖p ≤p

p− 1‖f‖p

which shows that the mapping f → F carries Lp into Lp.

9

(b) Prove that equality holds only if f = 0a.e.

(c) Prove that the constant p/(p− 1) cannot be replaced by a smaller one

.

(d) If f > 0 and f ∈ L1, prove that F /∈ L1.

[Proof]: (a) Firstly, assume that f ≥ 0 and f ∈ Cc((0,∞)). Then

F (x) =1

x

∫ x

0

f(t)dt (0 < x <∞)

is differential on (0,∞) and so xF ′(x) = f(x) − F (x), ∀x ∈ (0,∞). Therefore

∫ ∞

0F p(x)dx = F p(x)x |∞0 −p

∫ ∞

0F p−1(x)F ′(x)dx

= −p∫ ∞

0F p−1(x)xF ′(x)dx

= −p∫ ∞

0F p−1(x)(f(x) − F (x))dx

= −p∫ ∞

0F p−1(x)f(x)dx + p

∫ ∞

0F p(x)dx

Thus

(p−1)

∫ ∞

0

F p(x)dx = p

∫ ∞

0

F p−1(x)f(x)dx ≤ p(

∫ ∞

0

F (p−1)q(x)dx)1

q (

∫ ∞

0

f p(x)dx)1

p

and so

‖F‖p ≤p

p− 1‖f‖p.

Moreover, it is easy that

‖F‖p ≤p

p− 1‖f‖p, ∀f ∈ Cc((0,∞)).

Given f ∈ Lp((0,∞)), since Cc((0,∞)) is dense in Lp((0,∞)), there exist a

sequence fn ⊂ Cc((0,∞)) such that

‖fn − f‖p → 0, as n→ ∞.

That is, for any ε > 0, there exists N > 0 such that ‖fn − f‖p < ε if n > N .

It shows by the Jensen’s Inequality that

|Fn(x) − F (x)| ≤ 1

x

∫ x

0|fn(t) − f(t)|dt)

≤ 1

x(∫ x

0|fn(t) − f(t)|pdt)

1

p · x1

q

≤ x−1

p‖fn − f‖p

and thus

Fn(x) → F (x), n→ ∞, ∀x ∈ (0,∞).

10

Since

‖Fn‖p ≤p

p− 1‖fn‖p, n = 1, 2, · · ·

it shows that Fn ∈ Lp and Fn ⊂ Lp is Cauchy sequence. By the completion

of Lp, it exists G(x) ∈ Lp satisfying

‖Fn −G‖p → 0, n→ ∞

and so ‖Fn‖p → ‖G‖p < +∞. By the Fatou’s Lemma,

‖F‖p ≤ limn→∞

‖Fn‖p = ‖G‖p < +∞

whence F ∈ Lp. Moreover,

‖F‖p ≤ ‖G‖p = limn→∞

‖Fn‖p ≤ limn→∞

p

p− 1‖fn‖p =

p

p− 1‖f‖p.

Therefore, the map f → F is continuous from Lp(0,∞) to Lp(0,∞).

(b) Firstly, assume that f ≥ 0 and f ∈ Cc(0,∞). Then, by (a), we have

∫ ∞

0F p(x)dx = p

p−1

∫ ∞

0F p−1(x)f(x)dx

≤ p

p−1(∫ ∞

0F (p−1)q(x)dx)

1

q (∫ ∞

0f p(x)dx)

1

p

≤ p

p−1(∫ ∞

0F p(x)dx)

1

q (∫ ∞

0f p(x)dx)

1

p

and so

‖F‖p = p

p−1‖f‖p

⇔ ∃ α, β ∈ R and αβ 6= 0 satisfying αF p = βf p a.e.

⇔ αF = βf a.e.

⇔ f = 0 a.e.

.

Therefore, it shows the conclusion f = 0 a.e. if f ∈ Cc(0,∞) and ‖F‖p =p

p−1‖f‖p.

(c)

(d)Suppose that f > 0 and f ∈ L1, then G =∫ ∞

0fdµ < +∞ and so it

exists N > 0 such that ∫ x

0

f(t)dt >G

2

when x > N . Thus∫ ∞

0

F (x)dx ≥

∫ ∞

N

1

x

∫ x

0

f(t)dt ≥G

2

∫ ∞

N

1

xdx = +∞

11

and therefore F /∈ L1.

18. Let µ be a positive measure on X. A sequence fn of complex

measurable functions on X is said to converge in measure to the measurable

function f if to every ε > 0 there corresponds an N such that

µ(x : |fn(x) − f(x)| > ε) < ε

for all n > N . (This notion is of importance in probability theory.) Assume

µ(X) <∞ and prove the following statements:

(a) If fn(x) → f(x)a.e., then fn → f in measure.

(b) If fn ∈ Lp(µ) and ‖fn − f‖p → 0, then fn → f in measure; here

1 ≤ p ≤ ∞.

(c) If fn → f in measure, then fn has a subsequence which converges

to fa.e.

Investigate the converses of (a) and (b). What happen to (a), (b),and (c)

if µ(X) = ∞, for instance, if µ is Lebesgue measure on R1?

[Proof]: (a) Set E = x ∈ X : limn→∞

fn(x) 6= f(x), then µ(E) = 0. For

every ε > 0, we define

Ek = x ∈ X : |fn(x) − f(x)| > ε, k = 1, 2, · · · ;

Fn =∞⋃

k≥n

Ek, n = 1, 2, · · · ;F =∞⋂

n=1

Fn.

It is easy that F ⊂ E and so µ(F ) = 0. Since Fn is a sequence of decreasing

measurable sets and µ(X) < +∞, it shows that

limn→∞

µ(Fn) = µ(F ) = 0.

As En ⊂ Fn, n = 1, 2, · · ·, it is trival that

limn→∞

µ(En) = 0.

It is then proved that fn → f in measure.

(b) For every ε > 0, suppose that

Ek = x ∈ X : |fn(x) − f(x)| > ε, k = 1, 2, · · · .

Then it follows that for 1 ≤ p <∞,

εpµ(En) ≤

∫X

|fn − f |pdµ = ‖fn − f‖pp.

12

Since ‖fn − f‖p → 0, it shows that

limn→∞

µ(En) = 0.

It is then proved that fn → f in measure. If p = +∞ and ‖fn − f‖p → 0,

there exists a national N > 0 satisfying

‖fn − f‖p < ε, whenever n > N.

Hence it shows

µ(En) = 0, whenever n > N.

It is then proved that fn → f in measure.

(c)Suppose that fn → f in measure. For every national k,there exists

national nk such that

µ(Ek) <1

2k, k = 1, 2, · · · .

where Ek = x ∈ X : |fnk(x) − f(x)| > 1

2k.

Define

F =∞⋂

n=1

∞⋃k≥n

Ek

then we have

µ(F ) ≤ µ(∞⋃

k≥n

Ek) ≤∞∑

k=n

µ(Ek) =1

2n−1, n = 1, 2, · · · .

It is obvious that µ(F ) = 0. For any x ∈ X − F , it follows that there exists

national N such that

|fnk(x) − f(x)| ≤

1

2k, for each national k > N.

Therefore, we obtain

limk→∞

fnk(x) = f(x).

It then shows that fn has a subsequence which converges to f a.e.

20. Suppose φ is a real function on R such that

φ(

∫1

0

f(x)dx) ≤

∫1

0

φ(f)dx

for every real bounded measurable f . Prove that φ is then convex.

13

[Proof]: Assume that

x, y ∈ R; 0 < λ < 1.

Then

λx+ (1 − λ)y =

∫1

0

(xχ[0,λ](t) + yχ[λ,1])(t)dt.

Thusφ(λx+ (1 − λ)y) = φ(

∫1

0(xχ[0,λ](t) + yχ[λ,1])(t)dt)

≤∫

1

0φ (xχ[0,λ](t) + yχ[λ,1])(t)dt

= λφ(x) + (1 − λ)φ(y).

Therefore, φ is convex.

14

chapter four

In this set of exercises, H always denotes a Hilbert space.

1. If M is a closed subspace of H, prove that M = (M⊥)⊥. Is there a

similar true statement for subspaces M which are not necessarily closed?

[Proof]: It is easy M ⊂ (M⊥)⊥. on the other hand, for any z ∈ (M⊥)⊥,

it exist x ∈ M and y ∈ M⊥ with z = x + y when M is a closed subspace of

H. Thus,

0 = (z, y) = (x, y) + (y, y) = (y, y)

and so y = 0. Therefore, z = x ∈ M . It thus follows that M ⊃ (M⊥)⊥ and

whence M = (M⊥)⊥.

Next, assume that M is not closed subspace. Since M ⊂ (M⊥)⊥ and

(M⊥)⊥ is closed, it is obvious that M ⊂ (M⊥)⊥. since M ⊂ M , it shows

M⊥

⊂ M⊥. Moreover, (M⊥)⊥ ⊂ (M⊥)⊥. Again since M = (M

⊥)⊥, it is

proved that M = (M⊥)⊥.

2. Let xn : n = 1, 2, 3, · · · be a linearly independent set of vectors in

H. Show that the following construction yields an orthonormal set un such

that x1, x2, · · · , xN and u1, u2, · · · , uN have the same span for all N .

[Proof]: Let xn : n = 1, 2, 3, · · · be a linearly independent set of vectors

in H. Put

u1 =x1

‖x1‖; vn = xn −

n−1∑k=1

(xn, uk)uk; un =vn

‖vn‖; n = 2, 3, · · ·

Then it is easy that

(v2, u1) = 0; (u2, u1) = 0

and so (vn, uk) = 0, k = 1, 2, · · · , n− 1. Therefore, un is an orthonormal set.

Moreover,

spanxn = spanun.

3. Show that Lp(T ) is separable if 1 ≤ p < ∞, but that L∞(T ) is not

separable.

1

[Proof]: Firstly, assume that 1 ≤ p < ∞, then C(T ) is dense in Lp(T ).

Let P be the set of all trigonometric polynomials and P be the set of all

trigonometric polynomials with rational coefficients, it is obvious that P is

dense in C(T )and P is also. Since

‖f − g‖p ≤ ‖f − g‖∞, ∀f, g ∈ C(T )

and so it shows that P is countable dense subset in Lp(T ). This proves that

Lp(T ) is separable.

Next, we prove that L∞(T ) is not separable. Assume that L∞(T ) is

separable and let un is the countable dense subset of L∞(T ). Put

ft(s) = χ[0,t](s); 0 < t < 2π

and extend it to the real axis R with period 2π. Thus,

ft ∈ L∞(T ) and ‖ft‖∞ = 1.

Take 0 < ε < 1

2, according to the assumption, it exists some uk satisfying

U(uk, ε) ∩ un is infinite set.

Put ft, ft′ ∈ U(uk, ε); 0 < t < t′ ≤ 2π, then

1 = ‖ft − ft′‖∞ ≤ ‖ft − uk‖∞ + ‖ft′ − uk‖∞ < 2ε < 1

It is contradiction. Therefore, L∞(T ) is not separable.

4. Show that H is separable if and only if H contains a maximal orthonor-

mal system which is at most countable.

[Proof]: First, there exists a countable dense subset M of H if H is

separable. Let N is the maximal linearly independent subset of M . By the

conclusion of Exercise 2, it shows the existence of a maximal orthonormal

system of H which is at most separable.

Second, suppose that un is a countabe maximal orthonormal system of

H. Put M = spanun. If M 6= H, ther exists an element v ∈ H and v /∈ M .

Furthermore,

v =

∞∑n=1

(v, un)un ∈ M ; v − v 6= 0.

2

Let u0 = v−v

‖v−v‖, then

‖u0‖ = 1; (u0, un) = 0, n = 1, 2, · · · .

Therefore, u0, u1, u2, · · · ⊂ H is a orthonormal set but this contradict the

maximality of u1, u2, · · · ⊂ H. It thus follows that M = H. So H is

separable.

5. If M = x : Lx = 0, where L is a continuous linear functional on H,

prove that M⊥ is a vector space of dimension 1(unless M = H).

[Proof]: If L = 0, then M = H and M⊥ = 0. in the following, assume

that L 6= 0. Then it exists unique element y 6= 0 of H such that

Lx = (x, y); ∀x ∈ H.

Thus, y ⊥ M and so M = spany⊥. That is, M⊥ = spany since M is a

closed subspace by the continuity of L. This complete the proof.

7. Suppose that an is a sequence of positive numbers such that∑

anbn <

∞ whenever bn ≥ 0 and∑

b2n < ∞. Prove that

∑a2

n < ∞.

[Proof]: Assume that∞∑

n=1

a2n = +∞. For any k ∈ N, it then exists nk ∈ N

such thatn

k∑n=1

a2

n > k; 1 < n1 < n2 < n3 < · · ·

Hence

c2

k =

nk+1∑

nk+1

a2

n > 1; k = 1, 2, · · · .

Put

bn =

an

kck

; nk ≤ n < nk+1, k = 1, 2, · · ·

0; 1 ≤ n < n1

Then∞∑

n=1

b2

n =∞∑

k=1

nk+1∑

n=nk+1

a2n

k2c2k

=∞∑

k=1

1

k2< +∞

but∞∑

n=1

anbn =∞∑

k=1

nk+1∑

n=nk+1

a2n

kck

=∞∑

k=1

ck

k= +∞

Therefore, the assumption is error and∑

a2n < ∞.

3

9. If A ⊂ [0, 2π] and A is measurable, prove that

limn→∞

∫A

cos nxdx = limn→∞

∫A

sin nxdx = 0.

[Proof]: If A ⊂ [0, 2π] and A is measurable, then χA ∈ L2([0, 2π]). Since

L2([0, 2π]) is Hilbert space and eint : n ∈ Z is the maximal orthonormal set

of L2([0, 2π]), and so we have

∑n∈Z

|(χA, eint)|2 = ‖χA‖2 < +∞

Therefore, it shows that

|(χA, eint)| → 0, |n| → ∞

Furthermore, it follows that

limn→∞

∫A

cos nxdx = limn→∞

∫A

sin nxdx = 0.

16. If x0 ∈ H and M is a closed linear subspace of H, prove that

min‖x − x0‖ : x ∈ M = max|(x0, y)| : y ∈ M⊥, ‖y‖ = 1.

[Proof]: Suppose that P and Q are the projections on M , M⊥ respec-

tively. For x0 ∈ H, it shows that x0 = Px0 + Qx0 and

‖Qx0‖ = min‖x − x0‖ : x ∈ M.

On the other hand, for any y ∈ M⊥ with ‖y‖ = 1, then

(x0, y) = (Px0 + Qx0, y) = (Qx0, y)

and whence

|(x0, y)| = |(Qx0, y)| ≤ ‖Qx0‖ · ‖y‖ = ‖Qx0‖.

Since Qx0 ∈ M⊥, we may put y0 = Qx0

‖Qx0‖when Qx0 6= 0. Then (x0, y0) =

‖Qx0‖. Thus

max|(x0, y)| : y ∈ M⊥, ‖y‖ = 1 = ‖Qx0‖.

This complet the proof.

4

chapter five

2. Prove that the unit ball(open or closed) is convex in every normed

linear space.

[Proof]: Let X be a normed linear space and S be the open nuit ball of

X. Assume that 0 ≤ α, β ≤ 1 and α + β = 1. For any x, y ∈ S,

‖αx + βy‖ ≤ ‖αx‖ + ‖βy‖ = α‖x‖ + β‖y‖ < α + β = 1

and so αx + βy ∈ S. It shows that S is convex.

4. Let C be the space of all continuous functions on [0, 1], with the

supremum norm. Let M consist of all f ∈ C for which

∫ 1

2

0

f(t)dt −

∫1

1

2

f(t)dt = 1.

Prove that M is closed convex subset of C which contains no element of min-

imal norm.

[Proof]: By use of the Lebesgue dominated convergence theorem, it is

easy that M is closed convex set of C. Moreover, for any f ∈ M ,

1 =

∫ 1

2

0

f(t)dt −

∫1

1

2

f(t)dt ≤

∫1

0

|f(t)|dt = ‖f‖1 ≤ ‖f‖∞.

If f ∈ C and ‖f‖∞ = 1, it holds that∫

1

0

|f(t)|dt = 1;

∫1

0

(1 − |f(t)|)dt = 0; 1 − |f(t)| ≥ 0, ∀t ∈ [0, 1]

and so |f(t)| = 1, ∀t ∈ [0, 1]. But in this case, f /∈ M . Therefore, M is closed

convex subset of C which contains no element of minimal norm.

5. Let M be the set of all f ∈ L1([0, 1]), relative to Lebesgue measure,

such that ∫1

0

f(t)dt = 1.

Show that M is a closed convex subset of L1([0, 1]) which contains infinitely

many elements of minimal norm.

1

[Proof]: Assume that 0 ≤ α, β ≤ 1 and α + β = 1. For any f, g ∈ M ⊂

L1([0, 1]),

∫1

0

[αf + βg](t)dt = α

∫1

0

f(t)dt + β

∫1

0

g(t)dt = α + β = 1

whence αf + βg ∈ M . So M is convex. Next, given

fn ⊂ M ; and ‖fn − f‖1 → 0, n → ∞.

Then∫1

0

fn(t)dt −

∫1

0

f(t)dt =

∫1

0

≤

∫1

0

|fn(t) − f(t)|dt = ‖fn − f‖1 → 0

and so

limn→∞

∫1

0

fn(t)dt =

∫1

0

f(t).

It follows that f ∈ M since fn ⊂ M and so M is closed convex subset.

Moreover,

‖f‖1 ≥ 1, ∀f ∈ M.

Since

ntn−1 ∈ M, ∀n ∈ N; and ‖ntn−1‖1 = 1

it shows that M contains infinitely many elements of minimal norm.

6. Let f be a bounded linear functional on a subspace M of a Hilbert

space H. Prove that f has a unique norm-preserving extension to a bounded

linear functional on H, and that this extension vanishes on M⊥.

[Proof]: If f = 0, we only need to take the bounded linear functional

F = 0 on H. In the following, we assume f 6= 0. Since f is a bounded

linear functional on a subspace M , it is easy to extend f to a bounded linear

functional f on M and ‖f‖ = ‖f‖. In fact, for any x ∈ M , there exists a

sequence xn ⊂ M such that ‖xn−x‖ → 0 and then define f(x) = limn→∞

f(xn).

So we can define a linear functional of H such that

F (x) =

f(x); x ∈ M

0; x ∈ M⊥

and then it is proved that F is the norm-preserving linear extension on H of f

and this extension vanishes on M⊥. Moreover, for any norm-preserving linear

2

extension g on H of f , g |M= f . Since M⊥ = M⊥

and H = M ⊕M⊥, it shows

that it is unique that the norm-preserving extension of f to a bounded linear

functional on H which vanishes on M⊥.

8. Let X be a normed linear space, and let X∗ be its dual space, as defined

in Sec. 5.21, with the norm

‖f‖ = sup|f(x)| : ‖x‖ ≤ 1.

(a)Prove that X∗ is a Banach space.

(b) Prove that the mapping f → f(x) is, for each x ∈ X, a bounded linear

functional on X∗, of norm ‖x‖.

(c) Prove that ‖xn‖ is bounded if xn is a sequence in X such that f(xn)

is bounded for every f ∈ X∗.

[Proof]: (a) It is obvious that X∗ is a normed linear space. Given fn ⊂

X∗ is a cauchy sequence, then for any ε > 0, there is a N > 0 such that

‖fn − fm‖ < ε; ∀n, m > N

and so

|fn(x) − fm(x)| ≤ ‖fn − fm‖ · ‖x‖ < ‖x‖ · ε; ∀n, m > N, ∀x ∈ X.

Thus fn(x) ⊂ C is also a cauchy sequence and so we can define

f(x) = limn→∞

fn(x); ∀x ∈ X.

It is easy that f is a linear functional on X. Since ‖fn‖ is a cauchy sequence,

ther is M > 0 satisfying

‖fn‖ ≤ M, n = 1, 2, · · · .

Hence, for ε, N as above, it shows for n > N the following

|f(x)| ≤ |fn(x)| + ‖x‖ε ≤ ‖x‖(M + ε); ∀x ∈ X.

It forces

|f(x)| ≤ M‖x‖; ∀x ∈ X

and so ‖f‖ ≤ M . That is, f ∈ X∗. Therefore, X∗ is a Banach space.

3

(b) For each x ∈ X, define

λ(f) = f(x); ∀f ∈ X∗

Then for any f, g ∈ X∗ and α, β ∈ C,

Λ(αf + βg) = (αf + βg)(x) = αf(x) + βg(x) = αΛ(f) + βΛ(g)

and we see that Λ is a linear functional on X∗. Since

|Λ(f)| = |f(x)| ≤ ‖f‖ · ‖x‖; ∀f ∈ X∗

whence ‖Λ‖ ≤ ‖x‖. From the Hahn-Banach Theorem, there is g ∈ X∗ such

that g(x) = ‖x‖ and ‖g‖ = 1. It implies that ‖Λ‖ = ‖x‖.

(c) Suppose that xn ⊂ X such that f(xn) is bounded for every f ∈

X∗. By (a),(b) and the Banach-Steinhaus Theorem, it is easy to prove that

‖xn‖ is bounded.

9. Let c0, l1, l∞ be the Banach spaces consisting of all complex sequences

x = ξi, i = 1, 2, · · ·, define as follows:

x ∈ l1if and only if‖x‖1 =∑

|ξi| < ∞.

x ∈ L∞if and only if‖x‖∞ = sup |ξi| < ∞.

c0 is the subspace of l∞ consisting of all x ∈ l∞ for which ξi → 0 as i → ∞.

Prove the following four statements.

(a) If y = ηi ∈ l1 and Λx =∑

ξiηi for every x ∈ c0, then Λ is a bounded

linear functional on c0, and ‖Λ‖ = ‖y‖1. Moreover, every Λ ∈ (c0)∗ is obtained

in this way. In brief, (c0)∗ = l1.

(b) In the same sense, (l1)∗ = l∞.

(c) Every y ∈ l1 induces a bounded linear functional on l∞, as in (a). However,

this does not give all of (l∞)∗, since (l∞)∗ contains nontrivial functionals that

vanish on all of c0.

(d) c0 and l1 are separable but l∞ is not.

[Proof]: (a) For each y = ηi ∈ l1, we define

Λx =

∞∑i=1

ξiηi; ∀x = ξi ∈ c0.

4

Then the above series converges absolutely and so Λ is linear mapping. Since

|Λx| ≤ ‖x‖∞ · ‖y‖1

it follows that ‖Λ‖ ≤ ‖y‖1. Thus, Λ ∈ (c0)∗. On the other hand, put

ei =

i︷ ︸︸ ︷0, 0, · · · , 1, 0, 0, · · · , ; i = 1, 2, · · · .

It is obvious that ei ∈ c0 and ‖ei‖∞ = 1, i = 1, 2, · · ·. For any Λ ∈ (c0)∗,take

ηi = Λei, i = 1, 2, · · · .

Then

|ηi| = |Λei| ≤ ‖Λ‖ · ‖ei‖∞ = ‖Λ‖.

Let y = ηi and then y ∈ l∞ with ‖y‖∞ ≤ ‖Λ‖. Assume xn = ξni , where

ξni =

ηi

|ηi|; ηi 6= 0, i = 1, 2, · · · , n;

0; ηi = 0, or i > n.

then xn ∈ c0 with ‖xn‖∞ = 1, n = 1, 2, · · ·. Moreover,

Λxn =n∑

i=1

|ηi| ≤ ‖Λ‖ · ‖xn‖∞ = ‖Λ‖

whence∞∑i=1

|ηi| ≤ ‖Λ‖

Thus y ∈ l1 and ‖y‖1 ≤ ‖Λ‖. Since for any x = ξi ∈ c0, x =∞∑i=1

ξiei and

so Λx =∞∑i=1

ξiηi. From the proof above, it implies that ‖Λ‖ ≤ ‖y‖1 and so

‖Λ‖ = ‖y‖1. Therefore, (c0)∗ = l1.

(b) For each y = ηi ∈ l∞, define

Λx =

∞∑i=1

ξiηi; ∀x = ξi ∈ l1.

Then the above series converges absolutely and so Λ is linear mapping. Since

|Λx| ≤ ‖y‖∞ · ‖x‖1

5

it shows that ‖Λ‖ ≤ ‖y‖∞. Thus, Λ ∈ (l1)∗. On the other hand, suppose that

Λ ∈ (l1)∗. Put

ηi = Λei, i = 1, 2, · · · .

Then

|ηi| = |Λei| ≤ ‖Λ‖ · ‖ei‖1 = ‖Λ‖.

Let y = ηi and then y ∈ l∞ with ‖y‖∞ ≤ ‖Λ‖. Since for any x = ξi ∈ l1,

x =∞∑i=1

ξiei and so Λx =∞∑i=1

ξiηi. It implies that ‖Λ‖ ≤ ‖y‖∞ and so ‖Λ‖ =

‖y‖∞. Therefore, (l1)∗ = l∞.

(c) For each y = ηi ∈ l∞, define

Λx =

∞∑i=1

ξiηi; ∀x = ξi ∈ l∞.

Then the above series converges absolutely and so Λ is linear mapping. Since

|Λx| ≤ ‖x‖∞ · ‖y‖1

it shows that ‖Λ‖ ≤ ‖y‖1. Thus, Λ ∈ (l∞)∗. Since c0 ⊂ l∞ is closed subspace,

by the Hahn-Banach theorem, there is f ∈ (l∞)∗ such that f |c0= 0. By (a),

(c0)∗ = l1 and so f /∈ l1. Therefore, l1 6= (l∞)∗.

(d) Since

spaneic0

= c0; spaneil1

= l1

it is easy to prove that c0, l1 are separable.

10. If∑

αiξi converges for every sequence (ξi) such that ξi → 0 as i → ∞,

prove that∑

|αi| < ∞.

[Proof]: Put

x = αi; xn = α1, α2, · · · , αn, 0, 0, · · ·; ξ = ξn ∈ c0.

and define

ρ(ξ) =

∞∑i=1

αiξi; ρn(ξ) =

n∑i=1

αiξi.

Then ρ is linear and ρn(n ∈ N) is bounded linear functionals on c0. Moreover,

‖ρn‖ =n∑

i=1

|αi|, n ∈ N.

6

Since

limn→∞

ρn(ξ) = ρ(ξ); ∀ξ ∈ c0

exists, by the Banach-Steinhaus Theorem, there is M > 0 satisfying

‖ρn‖ ≤ M ; n ∈ N

That is,n∑

i=1

|αi| ≤ M, n ∈ N.

It thus forces that∑∞

i=1|αi| < +∞.

11. For 0 < α ≤ 1, let Lipα denote the space of all complex functions f

on [a, b] for which

Mf = sups6=t

|f(s) − f(t)|

|s − t|α< ∞.

Prove that Lipα is a Banach space, if ‖f‖ = |f(a)| + Mf ; also, if

‖f‖ = Mf + supx∈[a,b]

|f(x)|.

[Proof]: It is easy to prove that Lipα is a normed linear space with the

corresponding norm. Let fn be a cauchy sequence of Lipα.

(a) Assume

‖f‖ = |f(a)| + Mf .

There thus exists G > 0 such that Mfn≤ G. For any ε > 0, there is N > 0

such that

|fn(a) − fm(a)| < ε; Mfn−fm< ε

when n, m > N . For every s ∈ (a, b],

|fn(s)−fm(s)|

|s−a|α≤ |fn(a)−fm(a)|

|s−a|α+ |fn(s)−fm(s)−fn(a)+fm(a)|

|s−a|α

≤ Mfn−fm+ |fn(a)−fm(a)|

|s−a|α

It implies that fn(s) is cauchy sequence for each s ∈ [a, b]. Therefore, we

can define

limn→∞

fn(s) = f(s); ∀s ∈ [a, b].

Since for any s, t ∈ [a, b] and s 6= t, for each n ∈ N,

|f(s)−f(t)|

|s−t|α≤ |fn(s)−fn(t)|

|s−t|α+ |fn(s)−f(s)|

|s−t|α+ |fn(t)−f(t)|

|s−t|α

≤ |fn(s)−f(s)|

|s−t|α+ |fn(t)−f(t)|

|s−t|α+ G

7

Let n → ∞, then

sups6=t

|f(s) − f(t)|

|s − t|α≤ G < +∞.

That is, f ∈ Lipα and so Lipα is a Banach space.

(b) Assume

‖f‖ = Mf + supx∈[a,b]

|f(x)|.

There thus exists G > 0 such that Mfn≤ G. Put ‖f‖∞ = supx∈[a,b] |f(x)| and

so ‖fn‖∞ is cauchy sequence. Then it exists a complex function f on [a, b]

such that

supx∈[a,b]

|fn(x) − f(x)| → 0; n → 0.

Moreover, for any s, t ∈ [a, b] and s 6= t, for each n ∈ N,

|f(s)−f(t)|

|s−t|α≤ |fn(s)−fn(t)|

|s−t|α+ |fn(s)−f(s)|

|s−t|α+ |fn(t)−f(t)|

|s−t|α

≤ |fn(s)−f(s)|

|s−t|α+ |fn(t)−f(t)|

|s−t|α+ G

Let n → ∞, then

sups6=t

|f(s) − f(t)|

|s − t|α≤ G < +∞.

That is, f ∈ Lipα and so Lipα is a Banach space.

16. Suppose X and Y are Banach spaces, and suppose Λ is a linear

mapping of X into Y , with the following property: For every sequence xn

in X for which x = lim xn and y = lim Λxn exist, it is true that y = Λx. Prove

that Λ is continuous.

[Proof]: Provided that X, Y are Banach spaces. Define

X ⊕ Y = (x, y) | x ∈ X, y ∈ Y

The addition and scalar multiplication on X ⊕ Y are difined in the obvious

manner as follows:

(x1, y1)+(x2, y2) = (x1+x2, y1+y2); α(x, y) = (αx, αy); ∀x, xi ∈ X, y, yi ∈ Y, i = 1, 2.

If put

‖(x, y)‖ = ‖x‖ + ‖y‖; ∀x ∈ X, y ∈ Y

8

it then implies that X ⊕ Y is normed linear space. Given (xn, yn) ⊂ X ⊕ Y

is cauchy sequence, and so xn ⊂ X and yn ⊂ Y are also cauchy sequences.

Therefore, it exist x ∈ X and y ∈ Y such that

‖xn − x‖ → 0; ‖yn − y‖ → 0; if n → ∞.

Hence,

‖(xn, yn) − (x, y)‖ = ‖(xn − x, yn − y)‖ = ‖xn − x‖ + ‖yn − y‖ → 0; n → ∞.

This implies that X ⊕ Y is Banach space. Put

G = (x, Λx) | x ∈ X

it is easy that G is a subspace of X⊕Y . Assume that (xn, Λxn) ⊂ G is cauchy

sequence, and then xn ⊂ x and Λxn ⊂ Y are aslo cauchy sequences. So

there exist x ∈ X and y ∈ Y such that

‖xn − x‖ → 0; ‖Λxn − y‖ → 0; n → ∞.

According to the property of Λ, it holds that Λx = y and (x, Λx) ∈ G. It thus

shows that G is closed and so G is a Banach space. Take T (x, Λx) = x for

each x ∈ x, then T is linear mapping and

‖T (x, Λx)‖ = ‖x‖ ≤ ‖x‖ + ‖Λx‖ = ‖(x, Λx)‖; ∀x ∈ X

whence ‖T‖ ≤ 1. Moreover, T is injective and surjective bounded linear

mapping. By the open mapping theorem, it show that T−1 is continuous and

so there is M > 0 such that ‖T−1‖ ≤ M . Thus

‖x‖ + ‖Λx‖ = ‖(x, Λx)‖ = ‖T−1x‖ ≤ M‖x‖; ∀x ∈ X

and so

‖Λx‖ ≤ M‖x‖; ∀x ∈ X.

It follows that Λ is continuous.

17. If µ is a positive measure, each f ∈ L∞(µ) defines a pultiplication

operator Mf on L2(µ) into L2(µ), such that Mfg = fg. Prove that ‖Mf‖ ≤

‖f‖∞. For which measure µ is it true that ‖Mf‖ = ‖f‖∞ for all f ∈ L∞(µ)?

For which f ∈ L∞(µ) does Mf map L2(µ) onto L2(µ)?

9

[Proof]: For each f ∈ L∞(µ), define

Mfg = fg; ∀g ∈ L2(µ)

Then it is easy that Mf is a linear operator from L2(µ) into L2(µ). Further-

more,

‖Mfg‖2 = ‖fg‖2 ≤ ‖f‖∞ · ‖g‖2; ∀g ∈ L2(µ)

Thus it shows that ‖Mf‖ ≤ ‖f‖∞.

If f ∈ L∞(µ) with 1

f∈ L∞(µ), then Mf maps L2(µ) onto L2(µ).

18. Suppose Λn is a sequence of bounded linear transformations from

a normed linear space X to a Banach space Y , suppose ‖Λn‖ ≤ M < ∞ for

all n, and suppose there is a dense set E ⊂ X such that Λnx converges for

each x ∈ E. Prove that Λnx converges for each x ∈ X.

[Proof]: For every x ∈ X, there is a sequence xk ⊂ E such that

‖xk − x‖ → 0(k → ∞) since E ⊂ X is a dense subset. Moreover, Λnxk ⊂ Y

converges for each k ∈ N. It thus exists a sequence yk ⊂ Y such that

Λnxk → yk, n → ∞; k = 1, 2, · · · .

For any ε > 0, there exists N > 0 such that

‖xk − xk′‖ <ε

M; ∀k, k′ > N.

and so

‖yk − yk′‖ = limn→∞

‖Λn(xk − xk′)‖ ≤ limn→∞

‖Λn‖ · ‖xk − xk′‖ ≤ M ·ε

M= ε

It hence implies that yk ⊂ Y is a cauchy sequence. Since Y is a Banach

space, there is y ∈ Y with ‖yk − y‖ → 0. Therefore,

‖Λnx − y‖ = ‖Λnx − Λnxk + Λnxk − yk + yk − y‖

≤ ‖Λn‖ · ‖xk − x‖ + ‖Λnxk − yk‖ + ‖yk − y‖

≤ M · ‖xk − x‖ + ‖Λnxk − yk‖ + ‖yk − y‖

This implies that limn→∞

Λnx = y.

10

chapter six

2. Prove that the example given at the end of Sec. 6.10 has stated

properties.

[Proof]:

3. Prove that the vector space M(X) of all complex regular Borel mea-

sures on a locally compact hausdorff space X is a Banach space if ‖µ‖ = |µ|(X).

[Proof]: Firstly, assume µ, λ ∈ M(X) and α ∈ C. Then αµ ∈ M(X) and

|µ|, |λ| are regular Borel measures. For any measurable set E in X and ε > 0,

it exist compact sets K1, K2 ⊂ E and open sets G1, G2 ⊃ E satisfying:

|µ|(E) < |µ|(K1) +ε

2; |λ(E)| < |λ|(K2) +

ε

2

|µ|(G1) < |µ|(E) +ε

2; |λ|(G2) < |λ|(E) +

ε

2Take

K = K1 ∪ K2; g = G1 ∩ G2

it implies that Kis compact and G is open. Moreover,

(|µ| + |λ|)(E) < (|µ| + |λ|)(K) + ε; (|µ| + |λ|)(G) < (|µ| + |λ|)(E) + ε

and so |µ|+|λ| is regular Norel measure. Since |µ+λ| and |µ|+|λ| are bounded

positive measures and

|µ + λ| ≤ |µ| + |λ|

it shows that |µ + λ| is regular Borel measure. In fact, we have the following

conclusion:

Suppose that µ is bounded positive measure, then

µ is regular ⇔ for any measurable set E and ε > 0, there are compact subset

K ⊂ E and open set G ⊃ E such that

µ(E − K) < ε; µ(G − E) < ε.

Therefore, it shows that M(X) is a linear space. Define

‖µ‖ = |µ|(X), ∀µ ∈ M(X)

1

it is easy to show that ‖µ‖ is norm and M(X) is normed linear space. By

Theorem 6.19, it is obtained that M(X) ∼= (C0(X))∗. Since (C0(X))∗ is a

Banach space and so M(X) is also.

4. Suppose 1 ≤ p ≤ ∞, and q is the exponent conjugate to p. Suppose µ is

a positive σ-finite measure and g is a measurable function such that fg ∈ L1(µ)

for every f ∈ Lp(µ). Prove that then g ∈ Lq(µ).

[Proof]: Since µ is a positive σ-finite measure, there is a sequence of

disjoint measurable sets Xn of X such that

X =∞⋃

n=1

Xn; µ(Xn) < +∞, n = 1, 2, · · · .

It thus shows that

χXn∈ Lp(µ), n = 1, 2, · · · .

Therefore,

g ∈ L1(Xn, µ), n = 1, 2, · · · .

It implies that g is finite almost everywhere on X and so we may assume that

g is finite on X. Let

En = x ∈ X | |g(x)| ≤ n, n = 1, 2, · · · .

Then En is a sequence of increasing measurable sets and X =⋃∞

n=1En. Put

gn = χEn· g and then gn are measurable on X. Moreover,

(a) gn(x) → g(x); ∀x ∈ X.

(b) |gn(x)| ≤ n; ∀x ∈ X, n = 1, 2, · · ·.

(c) gn ∈ Lq(µ), n = 1, 2, · · ·.

Define

Tn(f) =

∫X

fgndµ; ∀f ∈ Lp(µ)

it is easy to prove that Tn is bounded linear functionals on Lp(µ) with

‖Tn‖ = ‖gn‖q; n = 1, 2, · · · .

Since

|Tn(f)| ≤

∫X

|fgn|dµ =

∫En

|fg|dµ ≤

∫X

|fg|dµ ≤ ‖fg‖1; n = 1, 2, · · ·

2

By use of the Banach-Steinhaus Theorem, there is M > 0 such that

‖Tn‖ ≤ M ; n = 1, 2, · · · .

It hence implies that

‖gn‖q ≤ M ; n = 1, 2, · · · .

If q = ∞, it shows from X =⋃∞

n=1En that there is n > 0 for any x ∈ X with

x ∈ En and so

|g(x)| = |gn(x)| ≤ ‖gn‖∞ ≤ M.

it shows that ‖g‖∞ ≤ M and g ∈ L∞(µ).

In the case 1 ≤ q < ∞, because

|gn(x)| ≤ |gn+1(x)|, n = 1, 2, · · · ; |gn(x)| → |g(x)|, ∀x ∈ X

By the Lebesgue’s monotone convergence Theorem, it shows∫

x

|gn|qdµ →

∫X

|g|qdµ, n → ∞

and so

(

∫X

|g|qdµ)1

q ≤ M.

Therefore, ‖g‖q ≤ M and g ∈ Lq(µ).

6. Sppose 1 < p < ∞ and prove that Lq(µ) is the dual space of Lp(µ)

even if µ is not σ-finite.

[Proof]: Sppose 1 < p < ∞ and µ is not σ-finite.

(a) Since for any f ∈ Lp(µ) and g ∈ Lq(µ), it is easy that f · g ∈ L1(µ). Then

we can define

Tg(f) =

∫x

f · gdµ, ∀f ∈ Lq(µ)

and Tg ∈ (Lp(µ))∗ with ‖Tg‖ ≤ ‖g‖q. Assume

En = x ∈ X : |g| ≥1

n, n = 1, 2, · · · .

It is obvious that En is an increasing sequence of measurable sets and

∞⋃n=1

En = E = x ∈ X : |g| > 0.

3

Put

fn = χEn|g|q−1α

where αg = |g|; |α| = 1. Then fn ∈ Lp(µ) and

‖fn‖p = (

∫En

|g|qdµ)1

p ; n = 1, 2, · · · .

Moreover, |fn| converges increasingly to |g|q−1. Therefore,

Tg(fn) =

∫X

fn · gdµ =

∫En

|g|qdµ; n = 1, 2, · · · .

and so ∫En

|g|qdµ ≤ ‖Tg‖ · ‖fn‖p = ‖Tg‖ · (

∫En

|g|qdµ)1

p

It shows that ∫En

|g|qdµ ≤ ‖Tg‖q; n = 1, 2, · · · .

By the Lebesgue’s monotone convergence Theorem, it shows that∫En

|g|qdµ =

∫X

|fn| · |g|dµ →

∫X

|g|qdµ; n → ∞.

Hence,∫

X|g|qdµ ≤ ‖Tg‖

q and so ‖g‖q = ‖Tg‖.

(b) Assume Φ ∈ (Lp(µ))∗ and ‖Φ‖ 6= 0. At first, for each f ∈ Lp(µ),

define

Df = x ∈ X : f(x) 6= 0; En = x ∈ X : |f(x)| ≥1

n.

Since Df =⋃∞

n=1En, and

µ(En)(1

n)p ≤

∫En

|f |pdµ ≤

∫X

|f |pdµ < +∞

it thus follows that µ(En) < +∞, n = 1, 2, · · ·; and so Df is σ-finite measur-

able set.

Secondly, given E ⊂ X is σ-finite measurable set, and put

M = f ∈ Lp(µ) : Df ⊂ E

then M ⊂ Lp(µ) is a Banach subspace. In fact, provided that fn ⊂ M is a

cauchy sequence, there is f ∈ Lp(µ) such that

‖fn − f‖p → 0, n → ∞.

4

So it exists a subsequence of fn ⊂ M which converges to f almost every-

where. It implies that f ∈ M . Therefore, Φ |M is a bounded linear functional

on M . Since it also that M ⊂ Lp(E, µ), by the Hahn-Banach Theorem, it can

extend Φ |M to a bounded linear functional Φ |E of Lp(E, µ) and so there is

unique hE ∈ Lq(E, µ) such that

Φ(f) =

∫E

f · hEdµ, ∀f ∈ M ; ‖hE‖q = ‖Φ |E ‖ = ‖Φ |M ‖ ≤ ‖Φ‖.

Take gE = hE on E and gE = 0 when x ∈ X − E. It is easy that

gE ∈ Lq(µ); DgE⊂ E; ‖gE‖q = ‖hE‖q ≤ ‖Φ‖

and

Φ(f) =

∫E

f · hEdµ =

∫X

f · gEdµ; ∀f ∈ M.

Order

α = sup‖gE‖q : E ⊂ X is σ − finite measurable subset

it is obvious that 0 ≤ α ≤ ‖Φ‖. Because there is a sequence fn ⊂ Lp(µ)

satisfying:

‖fn‖p = 1, n = 1, 2, · · · ; |Φ(fn)| → ‖Φ‖, n → ∞.

Take F =⋃∞

n=1Dfn

, then F is a σ-finite measurable set. Moreover,

|Φ(fn)| ≤ ‖Φ |F ‖ = ‖gF‖q ≤ α ≤ ‖Φ‖, n = 1, 2, · · · .

Therefore, it holds that ‖Φ‖ = ‖gF‖q = α. Write g = gF , then g ∈ Lq(µ) and

‖Φ‖ = ‖g‖q.

Thirdly, assume that A ⊂ X is σ-finite measurable set and write B = A ∪ F ,

B is also σ-finite measurable set. Then

‖Φ‖ = α = ‖gF‖q ≤ ‖gB‖q ≤ α ≤ ‖Φ‖.

That is,

‖Φ‖ = ‖gF‖q = ‖gB‖q.

Since Lp(F, µ) ⊂ Lp(B, µ), it show sthat hB = hF on F almost everywhere

5

and so gB = gF a.e. on F . Again since

‖Φ‖q =∫

X|gB|

qdµ

=∫

B|gB|

qdµ

=∫

A−F|gB|

qdµ +∫

F|gB|

qdµ

=∫

A−F|gB|

qdµ +∫

F|gF |

qdµ

≥∫

F|gF |

qdµ

= ‖Φ‖q

Hence, it holds that∫

A−F|gB|

qdµ = 0 and so gB = 0 a.e. on A− F . It implies

that g = gF = gB a.e. on X.

finally, from the conclusion above, it shows that

Φ(f) =

∫X

gDf∪Ffdµ =

∫X

gfdµ.

Therefore, (Lp(µ))∗ = Lq(µ).

6